Question

Question below:

Answer 1

what's the next step?

Answer 2

@SolomonZelman @Loser66 @freckles

Answer 3

\(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\frac{\sqrt{2x^2+x+1}}{3x-1}}\)

Answer 4

you can do this:

\(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\frac{\sqrt{2x^2+x+1\color{white}{\LARGE |}}}{3x-1}}\)


\(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\frac{\sqrt{2x^2+x+1\color{white}{\LARGE |}}}{\sqrt{(3x-1)^2}}}\)


\(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\sqrt{\frac{2x^2+x+1}{(3x-1)^2}}}\)

Answer 5

\(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\sqrt{\frac{2x^2+x+1}{9x^2-6x+1}}=\sqrt{2/9}=\sqrt{2}/3}\)

Answer 6

you can bring the limit inside the root, and do L'Hospital's rule twice.

Answer 7

how will i apply x approaches -infinity though?

Answer 8

Ok,

\(\large\color{black}{\displaystyle\lim_{x \to~ -\infty}\sqrt{\frac{2x^2+x+1}{9x^2-6x+1}}=\sqrt{\lim_{x \to~ -\infty}\frac{2x^2+x+1}{9x^2-6x+1}}}\)

Answer 9

\(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ -\infty}\frac{2x^2+x+1}{9x^2-6x+1}}}\)

You could apply L'Hospital's rule at this point,
because as x\(\to -\infty\), the top and bottom tend to ∞.
(an ∞/∞ case)

Answer 10

differentiate on top and bottom and you get

\(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ -\infty}\frac{4x+1}{18x-6}}}\)

Answer 11

but we haven't learned l'hospital's rule yet

Answer 12

Oh, so you aren't allowed to use it? (But do you know that rule, though?)

Answer 13

no, we're not allowed and I don't know that rule also

Answer 14

\(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ -\infty}\frac{2x^2+x+1}{9x^2-6x+1}}}\)

You can just conclude the answer based on the leading coefficients, then.
(The limit will be equal to 2/9 and the square root of 2/9 is going to be equal to √2 /3

Answer 15

It really did seem like an L'Hospital's rule kind of a problem, I apolgoize for mentioning it.

Answer 16

sorry i was off. I just finished my class.

will it always work though? for those kind of problems? Just get the leading coefficients?

I want to learn that rule.. where can I learn it? maybe I can just search it up on the internet.